Question

A quality controller at a beverage manufacturer is concerned that a bottling machine is under-filling (an opaque container) that is supposed to contain 1000 mL of fluid. A random sample of 20 containers is therefore taken and the volume in each recorded in the file STA201 201960 Assn Bottles.xlsx. In researching the problem the quality controller sees a statement in the machine’s manual that “volumes dispensed by the machine will follow a normal distribution”. (a) Based on the information provided, write down the null and alternate hypothesis that the quality controller should employ to test this concern. (b) Say why the statement in the manual is important to this analysis and then write down the name of the test to be employed. (c) Write down the decision rule in terms of a test statistic and give the corresponding decision rule using a p value. Note: Use a 5% level of significance for the test. (d) Calculate the value of the test statistic manually. (e) State whether you reject the null hypotheses or otherwise, justify your decision and then draw a conclusion that answers the original question.

Bottle	Volume
1	968.22
2	918.98
3	942.76
4	1024.02
5	988.96
6	1057.26
7	987.28
8	970.06
9	947.76
10	1003.18
11	1005.7
12	1076.16
13	931.36
14	990.06
15	950.64
16	1058.82
17	1036.26
18	928.64
19	898.16
20	978.54

Answer 1

From the given data we get the following summary statistics

	Mean (M)	Sample Size (n)	Standard Deviation (SD)
Population	1000
Sample	983.141	20	49.5874

a) Since we want to check if the plant is underfilling, that is filling less than 1000 mL,

the null and alternative hypotheses are          

Ho : μ = 1000       μ is the population mean volume of fluid in a bottle
Ha : μ < 1000           

b) The statement in the manual that “volumes dispensed by the machine will follow a normal distribution”.

is important, because we will be conducting a Student's t-test and the given statement of normality is the basic

assumption for a t-test

Since the population variance is unknown and the sample size is small, we use the t-test

c) Level of significance = 5% = 0.05

Critical value of t for 5% level of significance = t-crit

Degrees of freedom = n - 1 = 20 - 1 = 19

t-crit is found using Excel function T.INV

t-crit = T.INV(0.05, 19) = -1.729

Decision Rule in terms of test statistc t is

Reject Ho if calculated test statistic t < -1.729

Decision Rule in terms of p-value is

Reject Ho if p-value < 0.05

d) Using the below formula, we get the test statistic

t = -1.5205

e)

-1.5205 > -1.729

that is t statistic > t-crit

Hence, by the decision rule for test statistic, we DO NOT REJECT Ho

Conclusion :

There does not exist enough statistical evidence at α = 0.05 to show that the bottling machine is under filling.